ODE Background: Complex Variables (4A) Background : Complex Variables (4A) 6 Young Won Lim 12/21/15...

Young Won Lim12/21/15

ODE Background: Complex Variables (4A)


Copyright (c) 2011 - 2014 Young W. Lim.

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ODE Background :Complex Variables (4A)

3 Young Won Lim12/21/15

Complex Numbers



Complex Numbers

i = −1

i2 = −1

i3 = −i

i4 = 1

i

−1 1

−i

∗ i∗ i

∗ i∗ i

(3,2)

i2⋅i = −i

i2⋅i2 = +1

real

imaginary

3 + i2

(3,2) two real numbers2-d coordinate

3 + i2 one complex numberwith real part of 3and imaginary part of 2



Coordinate Systems

x = r cos

y = r sin

(r ,θ)

(x , y )

θ

r

(a) Cartesian Coordinate System

(b) Polar Coordinate System

r = √ x2 + y2

tanθ =yx



Coordinate Systems and Complex Numbers

real

imaginary

θ

r

(a) Cartesian Coordinate System

(b) Polar Coordinate System

θ

(d) Complex Number (II)

(c) Complex Number (I)

x + i y

r e iθ

(x , y )

(r ,θ)



Complex Numbers

real

imaginary

θ

(d) Complex Number (II)

(c) Complex Number (I)

x + i y

r e iθ

x = r cos

y = r sin

r = √ x2 + y2

tanθ =yx

= r (cosθ + i sinθ)

x + i y = r cosθ + i r sin θ

= r eiθ

real

imaginary

eiθ = cosθ + isin θ



Euler's Formula (1)

eiθ = cosθ + isin θ

ℜ{eiθ} = cosθ

ℑ{ei θ} = sinθ

i

−1 1

−i

∗ i∗ i

∗ i∗ i

e iπ/2

e iπ

e i3π/2

e0

x

y

= 0 + 1ie iπ/2

=−1 + 0 ie iπ

= 0 − 1ie i3π/2

= +1 + 0ie0

= +i = e−i3π/2

=−1 = e−iπ

= −i = e−iπ /2

=+1 = e−i2π



Euler's Formula (2)

e+iθ = cosθ + isin θ

e0

e iπ/4

e iπ/2

e i3 π/4

e iπ

e i5 π/4

e i3 π/2

e i7 π/4

e0

e−i7 π/4

e−i3π/2

e−i5 π/4

e−iπ

e−i3 π/4

e−π /2e−π /4

e−iθ = cosθ − isin θ



Absolute Values and Arguments

∣r eiθ∣ = ∣r∣∣eiθ∣ = r √cos2θ + sin2θ

arg (r eiθ)

r (cosθ + i sinθ)

∣z∣ = r

arg (z) = θ

absolute value

argument, phase

r cosθ

r sin θ

z = r eiθ =

r

cosθ

sinθ1



Computing Complex Number Argument

π2

−π2

π2

−π2

π

−π

yx> 0

yx< 0

yx< 0

yx> 0

atan( yx )atan( yx ) + π

atan( yx )atan( yx )− π

atan( yx ) + π

atan( yx )− π

atan( yx )atan(y/x) atan2(y, x)

[0, +π]

[0, −π]

+π−π



sin and cos

ei = cos i sin

ℜ{ei } = cos =

ei e−i

2

ℑ{ei } = sin =

ei − e−i

2 i

e−i = cos − i sin

e+iθ= cosθ + i sinθ

−e−iθ=−cosθ + i sinθ

e+iθ= cosθ + i sinθ

e−iθ= cosθ − i sinθ



y

z

y

z

x

x

z

y

x

Complex Exponential

real

imag

time

real

imag

time

real

imag

time

2π

e j t= cos t j sin t

phase

time

Top View

Front View

Side View

θ = ω t



ℜ

ℑ

t

ℜ

ℑ

t

e+ jω0t = cos (ω0 t) + j sin (ω0 t) e− jω t= cos (ω0 t) − j sin (ω0 t)

e+ j t= cos ( t) + j sin (t) (ω0 = 1) e− j t

= cos ( t) − j sin (t) (ω0 = 1)

Conjugate Complex Exponential



ℜ

ℑ

t

e+ j t= cos ( t) + j sin (t)

e− j t= cos ( t) − j sin (t)

(e+ j t+ e− j t

) = 2cos (t)

Cos(ω0t)

=A2

e+ jω0t +A2

e− jω0t

x (t) = A cos (ω0 t)



ℜ

ℑ

t

e+ j t= cos ( t) + j sin (t)

−e− j t= −cos (t) + j sin (t )

(e+ j t− e− j t

) = 2 j sin ( t)

x (t) = A sin (ω0 t)

Sin(ω0t)

=A2 j

e+ jω0 t −A2 j

e− jω0 t



Complex Power Series

e x= 1 x

x2

2!

x3

3!

x4

4!⋯

ei = 1 i

i2

2!i 3

3!i 4

4!i5

5!⋯

ei = cos i sin

= 1 i − 2

2!− i

3

3!

4

4! i

5

5!⋯

= 1− 2

2!

4

4!⋯ i −

3

3!

5

5!⋯

sin x = x −x3

3!

x5

5!−

x7

7!⋯

cos x = 1−x2

2!

x 4

4!−

x6

6!⋯



Taylor Series

f x = f a f ax−a f a

2!x−a2 ⋯

f nan!

x−an ⋯

x = a

f af ' af ' ' af 3 a⋮

f x



Maclaurin Series

f x = f a f ax−a f a

2!x−a2 ⋯

f nan!

x−an ⋯

x = 0

f 0 f ' 0 f ' ' 0 f 3 0 ⋮

f x

f x = f 0 f 0 xf 0

2!x2⋯

f n0n!

xn ⋯



Power Series Expansion

sin x = x −x3

3!

x5

5!−

x7

7!⋯

cos x = 1−x2

2!

x 4

4!−

x6

6!⋯

e x= 1 x

x2

2!

x3

3!

x4

4!⋯

f x = f 0 f 0 xf 0

2!x2⋯

f n0n!

xn ⋯



Complex Arithmetic

z = a + i b

w = c + i d

z+w = (a+c) + i(b+d)

z = a + i b

w = c + i d

z−w =(a−c) + i(b−d)

z = a + i b

w = c + i d

z w = (ac−bd ) + i(ad+bc)

z = a + i b

w = c + i d

zw=

= ( ac+bdc2+d2 ) + i(−ad+bc

c2+d2 )

= ( a + i bc + i d )(

c − i dc − i d )

(a + i bc + i d )



z = x + i y = ℜ{z}+ iℑ{z}

Complex Conjugate

z = x − i y = ℜ{z}− iℑ{z}

ℜ{z} =12(z + z)

ℑ{z} =12 i(z − z)

z + w = z + w

z − w = z − w

zw = z w

( zw ) =

zw

1z=

1z⋅zz=

zz z

=z

x2 + y2=

z

∣z∣2

z z = (x + i y)(x − i y) = x2+ y2



a = e loge a = e lna

Complex Power (1)

ab = (e logea)b = (eln a)b = eb lna

ai b= (eloge a)

i b= (eln a

)ib= eib lna

= cos(b lna) + i sin (b ln a)

ac+ib= ac

(eloge a)ib= ac

(eln a)i b= ac ei b ln a



Complex Power (2)

ai b= ei b ln a = [cos(b lna) + i sin(b lna)]

ac+ ib= ac ei b ln a

a = e lna

ab= eb ln a

= ac [cos(b lna) + i sin (b lna)]



Matrices



Determinant (1)

Determinant of order 2

[a1 a2

b1 b2] ∣a1 a2

b1 b2∣ = a1b2 − a2b1


[a1 a2 a3

b1 b2 b3

c1 c2 c3] [

+ − +− + −+ − + ]

[a1

b2 b3

c2 c3] [

a2

b1 b3

c1 c3] [

a3

b1 b2

c1 c2]



Determinant (2)


[a1 a2 a3

b1 b2 b3

c1 c2 c3] [

+ − +− + −+ − + ]

[a1

b2 b3

c2 c3] [

a2

b1 b3

c1 c3] [

a3

b1 b2

c1 c2]

∣a1 a2 a3

b1 b2 b3

c1 c2 c3∣ = + a1∣b2 b3

c2 c3∣ − a2∣b1 b3

c1 c3∣ + a3∣b1 b2

c1 c2∣

+ − +



Determinant (3)


[a1 a2 a3

b1 b2 b3

c1 c2 c3] [

+ − +− + −+ − + ]

[a2

b1 b3

c1 c3] [

a1 a3

b2

c1 c3] [

a1 a3

b1 b3

c2]

∣a1 a2 a3

b1 b2 b3

c1 c2 c3∣ = − a2∣b1 b3

c1 c3∣ + b2∣a1 a3

c1 c3∣ − c2∣a1 a3

b1 b3∣

−

+

−



Determinant – Rule of Sarrus


= a11∣a22 a23

a32 a33∣− a12∣a21 a23

a31 a33∣+ a13∣a21 a22

a31 a32∣[

a11 a12 a13

a21 a22 a23

a31 a32 a33] ∣

a11 a12 a13

a21 a22 a23

a31 a3 2 a33∣

[a11 a12 a13

a21 a22 a23

a31 a32 a33]

a11 a12 a13

a21 a22 a23

a31 a32 a33

= a11(a22 a33−a23 a32)

− a12(a21a33−a23a31)

+ a13(a21 a32−a22a31)

[a11 a12 a13

a21 a22 a23

a31 a32 a33]

a11 a12 a13

a21 a22 a23

a31 a32 a33[a11 a12 a13

a21 a22 a23

a31 a32 a33]

a11 a12 a13

a21 a22 a23

a31 a32 a33

Determinant of order 3 only

Recursive Method

Rule of Sarrus

+ − + − + −

+ a11a22a33− a11a23a32 + a12 a23 a31− a12a21a33 + a13 a21a32− a13 a22 a31



Solving Linear Equations

A set of linear equations

[a bc d ][ xy ] = [ef ]

a x + b y = e

c x + d y = f [ xy ] = [a bc d]

−1

[ef ]

∣a bc d∣ = ad − bc ≠ 0

∣e bf d∣

[ xy ] = 1ad − bc [ d −b

−c a ][ef ]

∣a ec f∣

x =∣e bf d∣∣a bc d∣

y =∣a ec f∣∣a bc d∣

= de − bf

= af − ce

[ xy ] = 1ad − bc [ de−bf

−ce+af ]

Cramer's Rule

If the inverse matrix exists



Cramer's Rule


a1 x + a2 y + a3 z = d

b1 x + b2 y + b3 z = e

c1 x + c2 y + c3 z = f [a1 a2 a3

b1 b2 b3

c1 c2 c3][

xyz ] = [

def ]

x =∣d a2 a3

e b2 b3

f c2 c3∣

∣a1 a2 a3

b1 b2 b3

c1 c2 c3∣

y =∣a1 d a3

b1 e b3

c1 f c3∣

∣a1 a2 a3

b1 b2 b3

c1 c2 c3∣

z =∣a1 a2 db1 b2 ec1 c2 f ∣∣a1 a2 a3

b1 b2 b3

c1 c2 c3∣



Linear Independence



Wronskian

W (f 1 , f 2 ,⋯, f n) = ∣f 1 f 2 ⋯ f n

f '1 f '2 ⋯ f 'n

⋮ ⋮ ⋯ ⋮

f 1(n−1) f 2

(n−1)⋯ f n

(n−1)∣

W (f 1(x) , f 2(x) , ⋯ , f n(x)) = ∣f 1(x) f 2(x) ⋯ f n(x)df 1

dx

df 2

dx⋯

df n

dx

⋮ ⋮ ⋯ ⋮

d(n−1) f 1

dx(n−1)

d(n−1) f 2

dx(n−1)⋯

d(n−1) f n

dx(n−1)

∣



Linear Independent Functions and Wronskian

C1= C2= 0y1 y2+ C2 = 0

always zero means all coefficients must be zero

C1

y1 and y

2 .are linearly independent functions

C1 y1 + C2 y2 = 0

C1 y1 ' + C2 y2 ' = 0 [ y1 y2

y1 ' y2 ' ][C1

C2] = [00 ]

[C1

C2] = [ y1 y2

y1 ' y2 ' ]−1

[00 ] = [00]∣ y1 y 2

y1 ' y2 ' ∣ ≠ 0

W ( y 1 , y 2) ≠ 0

the only solution:trivial

If the inverse matrix exists



Linear Dependent (1)

u

vw

u

vw

u

vw

w = u + v v = w − uu = w − v

{u , v , w} linearly dependent

u + v − w = 0 u + v − w = 0 u + v − w = 0

w in terms of u & v u in terms of w & v v in terms of w & u




u

vw

u

vw

u

vw

{u , v , w} linearly dependent

k 1u + k 2v + k 3 w = 0 m1u + m2 v + m3 w = 0 n1u + n2 v + n3 w = 0

(k1= 0)∧(k 2= 0)∧(k3 = 0 )(k1≠ 0)∨(k 2≠ 0)∨(k3 ≠ 0 )

(m1 = 0 )∧(m2 = 0 )∧(m3 = 0)(m1 ≠ 0 )∨(m2 ≠ 0 )∨(m3 ≠ 0)

(n1= 0)∧(n2 = 0 )∧(n3= 0)(n1≠ 0)∨(n2 ≠ 0 )∨(n3≠ 0)

w1 in terms of u & v w2 in terms of u & v w3 in terms of u & v




v4

v2v1

{v1 , v2 , v 3 , v 4} linearly dependent

k 1v1 + k 2 v2 + k3 v 3 + k 4 v 4 = 0

(k1 ≠ 0)∨(k2 ≠ 0)∨(k3 ≠ 0)∨(k 4 ≠ 0)

v3

m1v 1 + m2 v2 + m3 v 3 + m4 v 4 = 0

(m1 ≠ 0)∨(m2≠ 0)∨(m3 ≠ 0)∨(m4 ≠ 0 )0 v1 + m2 v2 + m3 v3 + m4 v4 = 0

(m1 = 0)∨(m2≠ 0)∨(m3 ≠ 0)∨(m4 ≠ 0)




v4

v2v1

{v1 , v2 , v 3 , v 4} linearly dependent

v3

=

=

=



Linear Independent (1)

S = {v1 , v2 , ⋯ , vn} V

k1v1 + k2v2 + ⋯ + knvn = 0

k1= k2= ⋯ = kn = 0

non-empty set of vectors in

the solution of the above equation

trivial solution:

v2

v1

−v3

if other solution exists linearly dependentS

if no other solution exists linearly independentS v1

v2

k 2 v2

k 1v1

k 3 v3v2

v1

v3v2

v1

v3v2



Linear Independent (2)

v1

v2

v1

v2

k1v1 + k2v2

every point in R2 can be represented by

linear combination of v1 and v2which are one set of linear independent two vectors

k1v1 + k2v2

only points on a line in R2

linear combination of v1 and v2which are one set of linear dependent two vectors



Basis

k1 e+ jθ

+ k2 e+ j θ

every complex number can be represented by

linear combination of and which are one set of linear independent two vectors

Basis : a set of linear independent spanning vectors

e+ jθ e+ jθe+ j θ

e− j θ

jsinθ

cosθ

e+ j θ

e− j θ

cosθ

j sinθl1 cosθ + l2 j sinθ

every complex number can also be represented by



Basis (2)

k1 e+ jθ

+ k2 e+ j θ

Basis : a set of linear independent spanning vectors

e+ j θ

e− j θ

jsinθ

cosθ

cosθ

j sinθ

l1 cosθ + l2 j sinθ

e+ j θ

e− j θ

k2e− jθ

k1e+ j θ

l2 j sinθ

l1 cosθ



Leibniz Formula



F (x)

Anti-derivative Examples

Anti-differentiation

f (x)

differentiation

Anti-derivative of f(x)

derivative of F(x)

f (x)=x2

=13

x3+ C∫ x2dxthe Indefinite

Integral of f(x)

∫0

x

f (x) dx = [ 13 x3]0

x

=13

x3

∫a

x

f (x) dx = [ 13 x3]a

x

=13

x3−

13a3

∫a

x

f (t ) dt = [ 13 t3]a

x

=13

x3−

13

a3

∫a

x

f (t ) dt =13

x3+ C

anti-derivative by the definite integral of f(x)

ddx∫a

x

f (t ) dt = f (x ) = x2



Fundamental Theorem of Calculus

F (x) = ∫a

x

f (t )dt

Derivative of an Antiderivative

F ' (x) = f (x )

∫a

b

f (t)dt = F (a)− F (b)

an Antiderivative and an Definite Integral

F ' (x) = f (x )

F (g(x )) = ∫a

g (x )

f (t)dt F ' (g (x)) = f (g(x))g ' (x)

f ( x)=x2F (x) = ∫a

x

t 2dt =13

x3−13

a3

F (2 x+1) = ∫a

2 x+1

t2dt =13(2 x+1)3−

13

a3 F ' (2 x+1) =dd x (

13(2 x+1)3−

13a3) = 1

3(2 x+1)2⋅2



Differentiation under the Integral Sign

dd x (∫a ( x)

b (x )

f (x , t)d t) = f (x , b(x))b ' (x) − f (x ,a(x))a ' (x) + ∫a (x )

b (x )

∂∂ x

f (x , t)d t

F (x) = ∫a

x

f (t )dt F ' (x) = f (x )

F (g(x )) = ∫a

g (x )

f (t)dt F ' (g (x)) = f (g(x))g ' (x)

dd x (∫a ( x)

b (x )

f (t )d t) = f (b(x))b ' (x) − f (a(x))a' (x)


References

[1] http://en.wikipedia.org/[2] M.L. Boas, “Mathematical Methods in the Physical Sciences”[3] E. Kreyszig, “Advanced Engineering Mathematics”[4] D. G. Zill, W. S. Wright, “Advanced Engineering Mathematics”[5] www.chem.arizona.edu/~salzmanr/480a

http://www.chem.arizona.edu/~salzmanr/480a

ODE Background: Complex Variables (4A) Background : Complex Variables (4A) 6 Young Won Lim 12/21/15...

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